TWO NEW TRIANGLES OF q-INTEGERS VIA q-EULERIAN POLYNOMIALS OF TYPE A AND B

GUONIU HAN, FRÉDÉRIC JOUHET AND JIANG ZENG

Abstract. The classical Eulerian polynomials can be expanded in the basis t^{k−1}(1 + t)^{n+1−2k} (1 ≤ k ≤ ⌊(n + 1)/2⌋) with positive integral coefficients. This formula implies both the symmetry and the unimodality of the Eulerian polynomials. In this paper, we prove a q-analogue of this expansion for Carlitz’s q-Eulerian polynomials as well as a similar formula for Chow-Gessel’s q-Eulerian polynomials of type B. We shall give some applications of these two formulae, which involve two new sequences of polynomials in the variable q with positive integral coefficients. An open problem is to give a combinatorial interpretation for these polynomials.

1. Introduction

The Eulerian polynomials A_{n}(t) := ∑_{k=1}^{n} A_{n,k}t^{k−1} (see [FS70, Fo09, St97]) may be defined by

∑_{k≥1} k^{n}t^{k} = \frac{A_{n}(t)}{(1 − t)^{n+1}} (n ∈ ℕ).

It is well known (see [FS70]) that there are nonnegative integers a_{n,k} such that

A_{n}(t) = ∑_{k=1}^{⌊(n+1)/2⌋} a_{n,k}t^{k−1}(1 + t)^{n+1−2k}.  (1.1)

For example, for n = 1, …, 4, the identity reads

A_{1}(t) = 1, A_{2}(t) = 1 + t, A_{3}(t) = (1 + t)^{2} + 2t^{2}, A_{4}(t) = (1 + t)^{3} + 8t(1 + t).

In particular, this formula implies both the symmetry and the unimodality (see for instance [Br08] for the definitions) of the Eulerian numbers (A_{n,k})_{1≤k≤n} for any fixed n. The coefficients a_{n,k} defined by (1.1) satisfy the following recurrence relation:

a_{n,k} = ka_{n−1,k} + 2(n + 2 − 2k)a_{n−1,k−1}  (1.2)

for n ≥ 2 and 1 ≤ k ≤ ⌊(n+1)/2⌋, with a_{1,1} = 1, and a_{n,k} = 0 for k ≤ 0 or k > ⌊(n+1)/2⌋.

\begin{matrix}n\k & 1 & 2 & 3 & 4 \\ 1 & 1 \\ 2 & 1 \\ 3 & 1 & 2 \\ 4 & 1 & 8 \\ 5 & 1 & 22 & 16 \\ 6 & 1 & 52 & 136 \end{matrix}  \begin{matrix}n\k & 0 & 1 & 2 & 3 \\ 1 & 1 \\ 2 & 1 & 4 \\ 3 & 1 & 20 \\ 4 & 1 & 72 & 80 \\ 5 & 1 & 232 & 976 \\ 6 & 1 & 716 & 766 & 3904 \end{matrix}

Table 1. The first values of (a_{n,k}) and (b_{n,k})

The classical Eulerian polynomials are the descent polynomials of the symmetric group or Coxeter group of type A. Analogues of Eulerian polynomials for other Coxeter groups were introduced and studied from a combinatorial point of view in the last three decades. Recall that for instance the Eulerian polynomials of type B are defined by

∑_{n≥0}(2k + 1)^{n}t^{n} = \frac{B_{n}(t)}{(1 − t)^{n+1}} .  (1.3)

The type B version of (1.1) appeared quite recently (see [Pe07,St08,Ch08]) and reads as follows

B_{n}(t) = ∑_{k=0}^{⌊n/2⌋} b_{n,k}t^{k}(1 + t)^{n−2k},  (1.4)

where b_{n,k} are positive integers satisfying the recurrence relation

b_{n,k} = (2k + 1)b_{n−1,k} + 4(n + 1 − 2k)b_{n−1,k−1},  (1.5)

for n ≥ 2 and 0 ≤ k ≤ ⌊n/2⌋, with b_{1,0} = 1, and b_{n,k} = 0 for k ≤ 0 or k > ⌊n/2⌋.

The numbers a_{n,k} and 4^{−k}b_{n,k} appear as A101280 and A008971 in The On-Line Encyclopedia of Integer Sequences : http://oeis.org.

The aim of this paper is to prove a q-analogue of (1.1) with a refinement of the triangle (a_{n,k}) for Carlitz’s q-Eulerian polynomials [Ca75], and also a q-analogue of (1.4) with a refinement of the triangle (b_{n,k}) for Chow-Gessel’s q-Eulerian polynomials of type B [CG07]. Note that some other extensions of (1.1) are discussed in [Br08,SW10,SZ10].

This paper is organized as follows: we derive in Section 2 a q-analogue of (1.1) using Carlitz’s q-Eulerian polynomials and derive some results about the q-tangent number T_{2n+1}(q) studied in [FH09]. In Section 3, we give a q-analogue of (1.4) using Chow-Gessel’s q-Eulerian polynomials of type B, which yields new q-analogues of the secant numbers. In Section 4, we apply our constructions to some conjectures on the unimodality from [CG07]. Finally, we will briefly give some concluding remarks in the fifth and last section.

TWO NEW TRIANGLES OF q-INTEGERS 3

2. A q-analogue for type A

The q-binomial coefficient \left[{n \atop k}\right]_{q} is defined by

\left[{n \atop k}\right]_{q} = \frac{(q; q)_{n}}{(q; q)_{k}(q; q)_{n−k}} , n ≥ k ≥ 0,

where (x; q)_{n} = (1−x)(1−xq) ⋯ (1−xq^{n−1}) and (x; q)_{0} = 1. Recall [Ca54] that Carlitz’s q-Eulerian polynomials A_{n}(t, q) := ∑_{k=1}^{n−1} A_{n,k}(q)t^{k} can be defined by

∑_{k≥0}[k + 1]_{q}^{n}t^{k} = \frac{A_{n}(t, q)}{(t; q)_{n+1}} ,  (2.1)

where [n]_{q} = 1 + q + ⋯ + q^{n−1}. It is easy to see that A_{n,k}(q) satisfy the recurrence:

A_{n,k}(q) = [k]_{q}A_{n−1,k}(q) + q^{k−1}[n + 1 − k]_{q}A_{n−1,k−1}(q) (1 ≤ k ≤ n).  (2.2)

The following is our q-analogue of (1.1).

Theorem 1. For any positive integer n, there are polynomials a_{n,k}(q) ∈ ℕ[q] such that the q-Eulerian polynomials A_{n}(t, q) can be written as follows:

A_{n}(t, q) = ∑_{k=1}^{⌊(n+1)/2⌋} a_{n,k}(q)t^{k−1}(−tq^{k}; q)_{n+1−2k}.  (2.3)

Moreover, the polynomials a_{n,k}(q) satisfy the following recurrence relation

a_{n,k}(q) = [k]_{q}a_{n−1,k}(q) + (1 + q^{k−1})q^{k−1}[n + 2 − 2k]_{q}a_{n−1,k−1}(q)  (2.4)

for n ≥ 2 and 1 ≤ k ≤ ⌊(n + 1)/2⌋, with a_{1,1}(q) = 1 and a_{n,k}(q) = 0 for k ≤ 0 or k > ⌊(n + 1)/2⌋.

Proof. Assume that a_{n,k}(q) are coefficients satisfying (2.4). Then, by the q-binomial formula (cf. [An98, Theorem 3.3]),

(z; q)_{N} = ∑_{j=0}^{N} \left[{N \atop j}\right]_{q}(−z)^{j}q^{j(j−1)/2},  (2.5)

we see that (2.3) is equivalent to:

A_{n,k}(q) = ∑_{s≥1} \left[{n + 1 − 2s \atop k − s}\right]_{q}q^{(k−s)s+\binom{k−s}{2}}a_{n,s}(q).  (2.6)

4 GUONIU HAN, FRÉDÉRIC JOUHET AND JIANG ZENG

Substituting (2.6) in (2.2), and using (2.4), we derive:

∑_{s≥1} \left[{n + 1 − 2s \atop k − s}\right]_{q}q^{(k−s)s+\binom{k−s}{2}} ([s]_{q}a_{n−1,s}(q) + (1 + q^{s−1})q^{s−1}[n + 2 − 2s]_{q}a_{n−1,s−1}(q))

= ∑_{s≥1} q^{(k−s)s+\binom{k−s}{2}} ([k]_{q}\left[{n − 2s \atop k − s}\right]_{q} + [n + 1 − k]_{q}\left[{n − 2s \atop k − 1 − s}\right]_{q}) a_{n−1,s}(q).

Extracting the coefficents of a_{n−1,s}(q) we obtain:

\left[{n + 1 − 2s \atop k − s}\right]_{q}[s]_{q} + \left[{n − 1 − 2s \atop k − s − 1}\right]_{q}(1 + q^{s})[n − 2s]_{q}

= [k]_{q}\left[{n − 2s \atop k − s}\right]_{q} + [n + 1 − k]_{q}\left[{n − 2s \atop k − 1 − s}\right]_{q}.

Canceling the common factors we get: [n + 1 − 2s]_{q}[s]_{q} + [n − k − s + 1]_{q}(1 + q^{s})[k − s]_{q} = [k]_{q}[n − k − s + 1]_{q} + [n + 1 − k]_{q}[k − s]_{q}. The last identity is easy to verify, and this shows that (2.3) is satisfied. □

The first values of the coefficients a_{n,k}(q) read as follows:

n\k 1 2 3
1 1
2 1
3 1  q + q^{2}
4 1  2q(1 + q)^{2}
5 1 q(1 + q)(3 + 5q + 3q^{2})  2q^{3}(1 + q)^{2}(1 + q^{2})
6 1 q(1 + q)^{2}(4 + 5q + 4q^{2}) q^{3}(1 + q)^{2}(1 + q^{2})(5 + 7q + 5q^{2})

In [FH09] Foata and Han defined a new sequence of q-tangent numbers T_{2n+1}(q) by

T_{2n+1}(q) = (−1)^{n}q^{\binom{n}{2}}A_{2n+1}(−q^{−n}, q).  (2.7)

We derive easily the following result from Theorem 1, which is the most difficult part of the main result in [FH09, Theorem 1.1]. Corollary 2. The q-tangent number T_{2n+1}(q) is a polynomial with positive integral coefficients. Proof. Let a_{n,k}^{∗}(q) = q^{−k(k+1)/2}a_{n,k}(q). Then (2.4) becomes

a_{n,k}^{∗}(q) = [k]_{q}a_{n−1,k}^{∗}(q) + (1 + q^{k−1})[n + 2 − 2k]_{q}a_{n−1,k−1}^{∗}(q)

with the same initial conditions as a_{n,k}(q). This proves that a_{n,k}^{∗}(q) is a polynomial in q with nonnegative integral coefficients. Now we show that T_{2n+1}(q) = a_{2n+1,n+1}^{∗}(q), which is sufficient to conclude. Replacing n by 2n + 1, k by n + 1, and t by −q^{−n} in (2.3), we get

A_{2n+1}(−q^{−n}, q) = ∑_{k=1}^{n+1} a_{2n+1,k}(q)(−q^{−n})^{k−1}(q^{k−n}; q)_{2n+2−2k} = a_{2n+1,n+1}(q)(−q^{−n})^{n},

since (q^{k−n}; q)_{2n+2−2k} = 0 for k = 1, 2, …, n. The result follows then from (2.7). □

We can also derive straightforwardly the following result, which was proved in [FH09] using combinatorics of the so-called doubloons.

Corollary 3. The quotient A_{2n}(t, q)/(1 + tq^{n}) is a polynomial in t and q with positive integral coefficients.

Proof. Note that

A_{2n}(t, q) = ∑_{k=1}^{n} a_{2n,k}(q)t^{k−1}(−tq^{k}; q)_{2n+1−2k}.

The result follows then from the fact that for k = 1, …, n, the coefficient (−tq^{k}; q)_{2n+1−2k} = (1 + tq^{k}) ⋯ (1 + tq^{2n−k}) contains the factor 1 + tq^{n}. □

For any nonnegative integer n, set

f_{n}(q) := ∑_{k=0}^{2n+1} \binom{2n + 1}{k}\frac{(−1)^{k}}{1 + q^{k−n}} .  (2.8)

Using the doubloon model, Foata-Han [FH09] proved that

d_{n}(q) := \frac{T_{2n+1}(q)}{(1 + q)(1 + q^{2}) … (1 + q^{n})} = \frac{(−1)^{n+1}(−1; q)_{n+2}}{(1 − q)^{2n+1}} f_{n}(q)

is a polynomial in ℕ[q]. Actually we can prove the integrality of d_{n}(q) without using the combinatorial device.

Proposition 4. We have d_{n}(q) ∈ ℤ[q].

Proof. Let g_{n}(q) = (−1)^{n+1}(−1; q)_{n+2}. Then f_{n}(q)g_{n}(q) is clearly a polynomial in ℤ[q]. We must show that 1 is a zero of order 2n + 1 of the polynomial f_{n}(q)g_{n}(q) or

d^{p}(f_{n}(q)g_{n}(q))/dq^{n}|_{q=1} = 0 for p = 0, …, 2n.

By Leibniz’s rule it suffices to show that f_{n}^{(p)}(1) = 0 for p = 0, …, 2n.

For any k ∈ ℤ and m ∈ ℕ, we define the Laurent polynomial P_{m,k}(x) by the relation:

h_{k}^{(m)}(x) = (\frac{d}{dx})^{m} (1 + x^{k})^{−1} = \frac{P_{m,k}(x)}{(1 + x^{k})^{m+1}} .

Thus P_{0,k} = 1, P_{1,k} = −kx^{k−1}, and for m ≥ 0, we have

P_{m+1,k}(x) = (1 + x^{k})P′_{m,k}(x) − k(m + 1)x^{k−1}P_{m,k}(x).

Therefore the P_{m,k} can, for m ≥ 1, be written as follows:

P_{m,k}(x) = ∑_{l=1}^{m} α_{l,m}x^{lk−m},

where α_{1,1} = −k and for m ≥ 1, α_{1,m+1} = (k − m)α_{1,m}, α_{m+1,m+1} = (m − k)α_{m,m},

α_{l,m+1} = (lk − m)α_{l,m} + (lk − mk − 2k − m)α_{l−1,m}, 2 ≤ l ≤ m.

This shows that for m ≥ 1 and 1 ≤ l ≤ m, the coefficient α_{l,m} is a polynomial in the variable k, with degree less than or equal to m. We deduce that P_{m,k}(1) = ∑_{l=1}^{m} α_{l,m} is also a polynomial in the variable k, with degree less than or equal to m, therefore we can write for some rational coefficients a_{j}(m) only depending on m:

h_{k}^{(m)}(1) = \frac{P_{m,k}(1)}{2^{m+1}} = ∑_{j=0}^{m} a_{j}(m)k^{j}.

Thus, differentiating (2.8) m times (m ≥ 0) and then setting q = 1, we get

f_{n}^{(m)}(1) = ∑_{k=0}^{2n+1} \binom{2n + 1}{k}(−1)^{k} ∑_{j=0}^{m} a_{j}(m)(k − n)^{j}

= ∑_{j=0}^{2n} a_{j}(m) ∑_{k=0}^{2n+1} \binom{2n + 1}{k}(−1)^{k}(k − n)^{j}.

Now, applying 2n + 1 times the finite difference operator Δ (defined by Δf(x) := f(x + 1) − f(x)) to the polynomial (n + 1 − x)^{j} (0 ≤ j ≤ 2n) and setting x = 0 we get

Δ^{2n+1}(n + 1 − x)^{j}|_{x=0} = ∑_{k=0}^{2n+1} \binom{2n + 1}{k}(−1)^{k}(k − n)^{j},

which should vanish because (n + 1 − x)^{j} is a polynomial in x of degree j < 2n + 1. □

3. A q-analogue for type B

A B_{n}-analogue of Carlitz’s q-Eulerian polynomials are introduced by Chow and Gessel [CG07]. These polynomials B(t, q) are defined by

∑_{k≥0}[2k + 1]_{q}^{n}t^{k} = \frac{B_{n}(t, q)}{(t; q^{2})_{n+1}} .  (3.1)

Let B(t, q) := ∑_{k=0}^{n} B_{n,k}(q)t^{k}. Then, the coefficients B_{n,k}(q) satisfy the recurrence relation [CG07, Prop. 3.2]:

B_{n,k}(q) = [2k + 1]_{q}B_{n−1,k}(q) + q^{2k−1}[2n − 2k + 1]_{q}B_{n−1,k−1}(q) 1 ≤ k ≤ n. (3.2)

We have the following B_{n}-analogue of (2.3).

Theorem 5. For any positive integer n, there are polynomials b_{n,k}(q) ∈ ℕ[q] such that the q-Eulerian polynomials of type B can be written as follows:

B_{n}(t, q) = ∑_{k=0}^{n} B_{n,k}(q)t^{k} = ∑_{k=0}^{⌊n/2⌋} b_{n,k}(q)t^{k}(−tq^{2k+1}; q^{2})_{n−2k}.  (3.3)

Moreover, the coefficients b_{n,k}(q) satisfy the following recurrence relation:

b_{n,k}(q) = [2k + 1]_{q}b_{n−1,k}(q) + (1 + q)(1 + q^{2k−1})q^{2k−1}[n + 1 − 2k]_{q^{2}}b_{n−1,k−1}(q) (3.4)

for n ≥ 2 and 0 ≤ k ≤ ⌊n/2⌋, with b_{1,0}(q) = 1, and b_{n,k}(q) = 0 for k < 0 or k > ⌊n/2⌋.

Proof. Assume that b_{n,k}(q) are coefficients satisfying (3.4). Then, by applying (2.5) with the substitution q ← q^{2}, we derive that (3.3) is equivalent to:

B_{n,k}(q) = ∑_{s≥0} \left[{n − 2s \atop k − s}\right]_{q^{2}}q^{k^{2}−s^{2}}b_{n,s}(q).  (3.5)

Substituting (3.5) in (3.2), and using (3.4), we get:

∑_{s≥0} \left[{n − 2s \atop k − s}\right]_{q^{2}}q^{k^{2}−s^{2}} ([2s + 1]_{q}b_{n−1,s}(q) + (1 + q)(1 + q^{2s−1})q^{2s−1}[n + 1 − 2s]_{q^{2}}b_{n−1,s−1}(q))

= ∑_{s≥0} q^{k^{2}−s^{2}} ([2k + 1]_{q}\left[{n − 1 − 2s \atop k − s}\right]_{q^{2}} + [2n + 1 − 2k]_{q}\left[{n − 1 − 2s \atop k − 1 − s}\right]_{q^{2}}) b_{n−1,s}(q).

Extracting the coefficents of b_{n−1,s}(q) we obtain:

\left[{n − 2s \atop k − s}\right]_{q^{2}}[2s + 1]_{q} + \left[{n − 2 − 2s \atop k − s − 1}\right]_{q^{2}}(1 + q)(1 + q^{2s+1})[n − 1 − 2s]_{q^{2}}

= [2k + 1]_{q}\left[{n − 1 − 2s \atop k − s}\right]_{q^{2}} + [2n + 1 − 2k]_{q}\left[{n − 1 − 2s \atop k − 1 − s}\right]_{q^{2}}.

Canceling the common factors yields:

[n − 2s]_{q^{2}}[2s + 1]_{q} + [n − k − s]_{q^{2}}(1 + q)(1 + q^{2s+1})[k − s]_{q^{2}}
= [2k + 1]_{q}[n − k − s]_{q^{2}} + [2n + 1 − 2k]_{q}[k − s]_{q^{2}}.

The last identity is easy to verify, and this proves (3.3). □

For n = 1, …, 4, equation (3.3) reads:

B_{1}(t, q) = 1 + qt;
B_{2}(t, q) = (−tq; q^{2})_{2} + (q + 2q^{2} + q^{3})t;
B_{3}(t, q) = (−tq; q^{2})_{3} + (2q + 5q^{2} + 6q^{3} + 5q^{4} + 2q^{5})t(1 + tq^{3});
B_{4}(t, q) = (−tq; q^{2})_{4} + (3q + 9q^{2} + 15q^{3} + 18q^{4} + 15q^{5} + 9q^{6} + 3q^{7})t(−tq^{3}; q^{2})_{2}

+ (2q^{4} + 7q^{5} + 11q^{6} + 13q^{7} + 14q^{8} + 13q^{9} + 11q^{10} + 7q^{11} + 2q^{12})t^{2}.

Theorem 5 implies immediately the following result, of which the first was derived in [FH10’, Theorem 1.1 (d)] with more work.

Corollary 6. For n ≥ 0, we have

B_{2n+1}(−q^{−2n−1}, q) = 0,  (3.6)
B_{2n}(−q^{−2n−1}, q) = (−1)^{n}q^{−n(2n+1)}b_{2n,n}(q).  (3.7)

Proof. By (3.3) we get

B_{2n+1}(−q^{−2n−1}, q) = ∑_{k=0}^{n} b_{2n+1,k}(q)(−q^{−2n−1})^{k}(q^{−2n+2k}; q^{2})_{2n+1−2k} = 0.

Substituting n by 2n and t by −q^{−2n−1} in (3.3) yields

B_{2n}(−q^{−2n−1}, q) = ∑_{k=0}^{n} b_{2n,k}(q)(−q^{−2n−1})^{k}(q^{−2n+2k}; q^{2})_{2n−2k}
= (−1)^{n}q^{−n(2n+1)}b_{2n,n}(q).  □

The above result leads to define a q-analogue of B_{2n}(−1) = (−1)^{n}4^{n}E_{2n} (where the E_{2n}’s are the famous secant numbers) by

E_{2n}^{∗}(q) := (−1)^{n}q^{n(n+1)}B_{2n}(−q^{−2n−1}, q).  (3.8)

Theorem 7. There is a polynomial G_{2n}^{∗}(q) ∈ ℤ[q] such that G_{2n}^{∗}(1) = E_{2n} and

E_{2n}^{∗}(q) = (1 + q)(1 + q^{3})(1 + q^{5}) ⋯ (1 + q^{2n−1}) · (1 + q)^{n} · G_{2n}^{∗}(q).

Proof. Recall that E_{2n}^{∗}(q) = (−1)^{n}q^{n(n+1)}B_{2n}(−q^{−2n−1}, q). From (3.1) we derive

\frac{B_{n}(t, q)}{(t; q^{2})_{n+1}} = (1 − q)^{−n} ∑_{j≥0}(1 − q^{2j+1})^{n}t^{j}

= (1 − q)^{−n} ∑_{j≥0} t^{j} ∑_{k=0}^{n} \binom{n}{k}(−q^{2j+1})^{k}

= (1 − q)^{−n} ∑_{k=0}^{n} \binom{n}{k}\frac{(−q)^{k}}{1 − tq^{2k}} .

Substituting n by 2n and setting t = −q^{−2n−1} we obtain

E_{2n}^{∗}(q) = (−1)^{n}q^{n(n+1)} \frac{(−q^{−2n−1}; q^{2})_{2n+1}}{(1 − q)^{2n}} ∑_{k=0}^{2n} \binom{2n}{k}\frac{(−q)^{k}}{1 + q^{2k−2n−1}} .

Let

G_{2n}^{∗}(q) := \frac{E_{2n}^{∗}(q)}{(1 + q)(1 + q^{3}) … (1 + q^{2n−1})(1 + q)^{n}}

= (−1)^{n}q^{−n−1} \frac{(−q; q^{2})_{n+1}}{(1 + q)^{n}(1 − q)^{2n}} ∑_{k=0}^{2n} \binom{2n}{k}\frac{(−q)^{k}}{1 + q^{2k−2n−1}} .

For any nonnegative integer n, set

f_{n}^{∗}(q) := ∑_{k=0}^{2n} \binom{2n}{k}\frac{(−q)^{k}}{1 + q^{2k−2n−1}} .  (3.9)

Let g_{n}^{∗}(q) = (−1)^{n}q^{−n−1}(−q; q^{2})_{n+1}/(1 + q)^{n}. Then f_{n}^{∗}(q)g_{n}^{∗}(q) is clearly a polynomial in ℤ[q]. We must show that 1 is a zero of order 2n of the polynomial f_{n}^{∗}(q)g_{n}^{∗}(q) or

d^{p}(f_{n}^{∗}(q)g_{n}^{∗}(q))/dq^{n}|_{q=1} = 0 for p = 0, …, 2n − 1.

By Leibniz’s rule it suffices to show that d^{p}(f_{n}^{∗}(q))/dq^{p}|_{q=1} = 0 for p = 0, …, 2n − 1. The rest of the proof is almost the same as that of Proposition 4, and is left to the reader. □

Conjecture 8. All the coefficients of the polynomials G_{2n}^{∗}(q) are positive.

Since G_{2n}^{∗}(1) = E_{2n}, the above conjecture would yield a new refinement of the secant number.

4. Application to unimodal problems

A sequence {α_{0}, …, α_{d}} is unimodal if there exists an index 0 ≤ j ≤ d such that α_{i} ≤ α_{i+1} for i = 1, …, j − 1 and α_{i} ≥ α_{i+1} for i = j, …, d. Chow and Gessel [CG07] studied a kind of unimodality property of the q-Eulerian numbers assuming that q is a real number. In this section, we derive some unimodal properties of the sequence (A_{n,k}(q))_{1≤k≤n} and (B_{n,k}(q))_{1≤k≤n} from our previous results. From Theorem 1, we are able to deduce the following corollary, which provides a further support to Conjecture 4.8 in [CG07].

Proposition 9. Let n ≥ 2 be an integer and j = ⌊(n + 1)/2⌋. Then for k = 1, …, j − 1, we have A_{n,k+1}(q) > A_{n,k}(q) if q > 1 and A_{n,n−k+1}(q) < A_{n,n−k}(q) if q < 1.

Proof. We start from (2.6), which can be rewritten

A_{n,k}(q) = ∑_{s=1}^{k} \left[{n + 1 − 2s \atop k − s}\right]_{q}q^{(k−s)(k+s−1)/2}a_{n,s}(q),

for k = 1, …, n, where we assume a_{n,s}(q) = 0 for s > j. Thus we can write for k = 1, …, j − 1:

A_{n,k+1}(q) − A_{n,k}(q) = a_{n,k+1}(q)

+ ∑_{s=1}^{k} \left[{n + 1 − 2s \atop k + 1 − s}\right]_{q}q^{(k+1−s)(k+s)/2}a_{n,s}(q) (1 − q^{−k} \frac{1 − q^{k+1−s}}{1 − q^{n+1−k−s}} ) .

We know that the q-binomial coefficient is a polynomial in q with nonnegative integer coefficients, and from Theorem 1 that this is also true for a_{n,s}(q), s = 1, …, k + 1. Therefore it is enough to show that the coefficient between brackets is nonegative for 1 ≤ s ≤ k ≤ j − 1. This coefficient can be rewritten as:

\frac{q^{n+1} − q^{k+s} + q^{s} − q^{k+1}}{q^{n+1} − q^{k+s}} .

Assume first that q > 1. As k+s ≤ 2j−2 ≤ n−1 < n+1, the denominator of this fraction is positive. Moreover, it is not difficult to see that under the conditions 1 ≤ s ≤ k ≤ j −1, and by using (n − 1)/2 ≤ j ≤ (n + 1)/2, we have the following inequalities:

q^{n+1} − q^{k+s} + q^{s} − q^{k+1} ≥ q^{n+1} − q^{2k} + q^{k} − q^{k+1}
≥ q^{n+1} − q^{2j−2} + q^{j−1} − q^{j}
≥ q^{n+1} − q^{n−1} + q^{(n−3)/2} − q^{(n+1)/2}.

This last expression can be rewritten (q^{(n+1)/2} − 1)(q^{(n+1)/2} − q^{(n−3)/2}) and is nonnegative, which shows that A_{n,k+1}(q) ≥ A_{n,k}(q) for k = 1, …, j − 1.

In the case 0 < q < 1, we only need to use the well-known relation A_{n,n−k+1}(q) = q^{n(n−1)/2}A_{n,k}(1/q) for any k = 1, …, n, and the result is obvious from the case q > 1. □

In the type B case, it is conjectured in [CG07, Conjecture 4.6] that the sequence (B_{n,k}(q))_{0≤k≤n} is unimodal. By Theorem 5, we are able to confirm partially this conjecture.

Proposition 10. Let n ≥ 2 be an integer and j = ⌊n/2⌋. Then for k = 1, …, j − 1, we have B_{n,k+1}(q) > B_{n,k}(q) if q > 1 and B_{n,n−k}(q) < B_{n,n−k−1}(q) if q < 1.

Proof. We start from (3.5), which can be rewritten

B_{n,k}(q) = ∑_{s=0}^{k} \left[{n − 2s \atop k − s}\right]_{q^{2}}q^{k^{2}−s^{2}}b_{n,s}(q),

for k = 0, …, n, where we assume b_{n,s}(q) = 0 for s > j. Thus we can write for k = 0, …, j − 1:

B_{n,k+1}(q) − B_{n,k}(q) = b_{n,k+1}(q)

+ ∑_{s=0}^{k} \left[{n − 2s \atop k + 1 − s}\right]_{q^{2}}q^{(k+1)^{2}−s^{2}}b_{n,s}(q) (1 − q^{−2k−1} \frac{1 − q^{2(k+1−s)}}{1 − q^{2(n−k−s)}} ) .

We know that the q-binomial coefficient is a polynomial in q with nonnegative integer coefficients, and from Theorem 5 that this is also true for b_{n,s}(q), s = 0, …, k + 1. Therefore it is enough to show that the coefficient between brackets is nonegative for 0 ≤ s ≤ k ≤ j − 1. This coefficient can be rewritten as:

\frac{q^{2n} − q^{2s+2k} + q^{2s−1} − q^{2k+1}}{q^{2n} − q^{2k+2s}} .

Assume first that q > 1. As k + s ≤ 2j − 2 ≤ n − 2 < n, the denominator of this fraction is positive. Moreover, it is not difficult to see that under the conditions 0 ≤ s ≤ k ≤ j −1, and by using n/2 − 1 ≤ j ≤ n/2, we have the following inequalities:

q^{2n} − q^{2s+2k} + q^{2s−1} − q^{2k+1} ≥ q^{2n} − q^{4k} + q^{2k−1} − q^{2k+1}
≥ q^{2n} − q^{4j−4} + q^{2j−3} − q^{2j−1}
≥ q^{2n} − q^{2n−4} + q^{n−5} − q^{n−1}.

This last expression can be rewritten (q^{2n} − q^{n−1})(1 − q^{−4}) and is nonnegative, which shows that B_{n,k+1}(q) ≥ B_{n,k}(q) for k = 0, …, j − 1.

In the case 0 < q < 1, we only need to use the well-known relation B_{n,n−k}(q) = q^{n^{2}}B_{n,k}(1/q) for any k = 0, …, n, and the result is obvious from the case q > 1. □

5. An open problem on the combinatorial interpretations

By Theorems 1 and 5, the polynomials a_{n,k}(q) and b_{n,k}(q) have positive integral coefficients. It is then natural to ask the following question.

Problem 11. What are the combinatorial interpretations for a_{n,k}(q) and b_{n,k}(q)?

We can give a combinatorial interpretation for the odd central terms a_{2n+1,n+1}(q) by using the doubloon model. Recall [FH09] that a doubloon of order (2n+1) is defined to be a permutation of the word 012 ⋯ (2n+1), represented as a 2×(n+1)-matrix δ = \binom{a_{0} ··· a_{n}}{b_{0} ··· b_{n}}. Define

cmaj′ δ := maj(a_{0} ⋯ a_{n}b_{n} ⋯ b_{0}) − (n + 1)des(a_{0} ⋯ a_{n}b_{n} ⋯ b_{0}) + n^{2},

where “des” and “maj” are the usual number of descents and major index defined for words. A doubloon δ = \binom{a_{0} ··· a_{n}}{b_{0} ··· b_{n}} is said to be interlaced, if for every k = 1, 2, …, n the sequence (a_{k−1}, a_{k}, b_{k−1}, b_{k}) or one of its three cyclic rearrangements is monotonic increasing or decreasing. By Theorem 1.5 in [FH09] we have the following result.

Proposition 12. The polynomial a_{2n+1,n+1}(q) is the generating function for the set of interlaced doubloons of order 2n + 1 by the statistic cmaj′.

Another sequence of q-secant numbers is introduced in [FH10’] by

E_{2n}(q) = (−1)^{n}q^{n^{2}}B_{2n}(−q^{−2n}, q).

Unfortunately, it seems not easy to relate our coefficients b_{n,k}(q) from Section 3 to the doubloons of type B, even for the central cases.

Acknowledgement

This work was partially supported by the grant ANR-08-BLAN-0243-03. The second author is grateful to Frédéric Chapoton for several discussions at the initial stage of this work.

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Guoniu Han, Institut de Recherche Mathématique Avancée, Université de Strasbourg et CNRS, 7, rue René-Descartes, F-67084 Strasbourg, France.

http://www-irma.u-strasbg.fr/~guoniu/

Frédéric Jouhet, Université de Lyon, Université Lyon I, CNRS, UMR 5208 Institut Camille Jordan, 43, bd du 11 Novembre 1918, 69622 Villeurbanne Cedex, France.

http://math.univ-lyon1.fr/~jouhet/

Jiang Zeng, Université de Lyon, Université Lyon I, CNRS, UMR 5208 Institut Camille Jordan, 43, bd du 11 Novembre 1918, 69622 Villeurbanne Cedex, France.

http://math.univ-lyon1.fr/~zeng/